Chemistry 332 - Winter 1997

Chapter 14 - Carbohydrates

Practice Exercises

Structure

D-Allose is the C-3 epimer of D-glucose. Write its structure in an open-chain form, in a Haworth projection (a beta-pyranose), and in a chair form.

Draw D-ribose in its pyranose form, in both a Haworth projection and in a chair form.

Draw the disaccharide that would result from linking D-glucose from its alpha-1 position to the C-4 position of D-galactose (which is the C-4 epimer of D-glucose).


Reactions

Identify the sugars above as reducing or non-reducing sugars.

Write the steps involved in converting open-chain D-ribose into methyl beta-D-ribofuranoside. Use methanol with acid catalysis.


The Nobel Prize in 1902

* Emil Fischer identifies the configuration of glucose *

This sequence of experiments illustrates the logic behind "how do we know" the structures of compounds. In the last 95 years, our experiments have become more complex and sophisticated but the art of creating informative experiments and the logic applied in their interpretation remain as valid as ever. Try to follow the award-winning logic.

Starting points:

a) Glucose was known to be an aldohexose.

b) All stereochemistry was related to the configuration at C-5
(Fischer assumed the C-5 OH was to the right - D - which was correct)

The Kiliani-Fischer synthesis was developed to convert an aldose to a mixture of the two epimers that have one more carbon:

Fischer's Experiments:

1) Arabinose, an aldopentose, leads to glucose and mannose.
(Conclusion: glucose and mannose are C-2 epimers, with their C-3,4,5 like arabinose)

2) Arabinose is oxidized by HNO3 to give an optically active aldaric acid.
(Conclusion: if C-5 is right, C-3 must be left - both right would have given an optically inactive (meso) aldaric acid)

3) The aldaric acids from glucose and mannose are both optically active.
(Conclusion: C-4 must be right - if C-4 were left, one of the two epimers would have given an optically inactive (meso) aldaric acid)

4) There is another sugar (gulose, which Fischer had to make) that gives the same aldaric acid as glucose.
(Conclusion: C-2 must be right for glucose and left for mannose - mannose gives an aldaric acid that cannot come from any other sugar)

Summary:

Glucose must have its OH groups oriented:

C-2 right
C-3 left
C-4 right
C-5 right